Question

prove that square of any odd integer is of the form of 8p + 1 where p is an integer​

Answer 1

Let us assume :-

Odd integer is of the form 4p + 1 and 4p + 3.

Suppose a be any positive integer

First case we have :-

⇒ a = 4p + 1

Squarring Both sides we get :-

⇒ a^2 = (4p + 1)^2

= 16p^2 + 8p + 1

= 8(2p^2 + q) + 1

⇒ p = (2p^2 + q)

= 8p + 1

Second Case we have :-

⇒ a = 4p + 3

Squarring Both sides we get :-

⇒ a^2 = (4p + 3)^2

= 16p^2 + 24p + 9

= 8(2p^2 + 3p + 1) + 1

⇒ p = (2p^2 + 3q + 1)

= 8p + 1

Square of any positive odd integer is of the form 8p + 1